December 13, 2019, 07:40:20 AM
Forum Rules: Read This Before Posting


Topic: Question about Density  (Read 605 times)

0 Members and 1 Guest are viewing this topic.

Offline Devabhai

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
Question about Density
« on: November 20, 2019, 01:20:20 AM »
Hello everyone,,
I know that this is probably going to seem SUPER simple to most of you out there, but for some reason I just can't wrap my mind around the concept of density.

So, density is equal to mass / volume. So if I'm trying to find the density of say hydrogen gas (H2) in STP conditions I can use the ideal gas law to solve for V:

PV = nRT

1 atm * V = 1 mol * 0.08206 atm(L) / mol(K) * 273,15 K

so V = 22,4 L

but I don't understand why we also can't use this equation to determine the number of moles in a 1 liter container:

1 atm * 1 L = n * 0.08206 atm(L) / mol(K) * 273,15 K

in which case n = 22,4 moles (which would mean that the density of hydrogen gas is equal to 45,2 grams per L)...

I know this is grade-school stuff, but for some reason I'm having difficulties processing this...Any help is greatly appreciated! :)

Offline chenbeier

  • Full Member
  • ****
  • Posts: 693
  • Mole Snacks: +61/-17
  • Gender: Male
Re: Question about Density
« Reply #1 on: November 20, 2019, 01:36:51 AM »
Who says that  you can't?

But your second calculation is wrong. Check it again.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 25365
  • Mole Snacks: +1663/-398
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Question about Density
« Reply #2 on: November 20, 2019, 03:13:23 AM »
So, density is equal to mass / volume. So if I'm trying to find the density of say hydrogen gas (H2) in STP conditions I can use the ideal gas law to solve for V:

PV = nRT

1 atm * V = 1 mol * 0.08206 atm(L) / mol(K) * 273,15 K

so V = 22,4 L

You were OK up to this moment. Now you know the volume of 1 mole of hydrogen. What is its mass? Just plug these numbers into the density definition.

Quote
1 atm * 1 L = n * 0.08206 atm(L) / mol(K) * 273,15 K

in which case n = 22,4 moles

No, while the first equation is OK, the result of your calculations of n is wrong. As chenbeier said, check your math.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 6916
  • Mole Snacks: +492/-81
  • Gender: Male
Re: Question about Density
« Reply #3 on: November 20, 2019, 03:36:11 AM »
The thing is very simple indeed. Each student rather knows the volume of 1 mole of gases V under STP conditions. It can also determine the molar mass of gas M from the periodic table. Density d = m / V
You counted V on a very complicated path. You can then calculate the density using this formula.
AWK

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 3206
  • Mole Snacks: +278/-57
Re: Question about Density
« Reply #4 on: November 21, 2019, 06:23:19 AM »
Somewhere a multiplication and a division were mixed up. If a mole takes more than a litre at STP, then you have less than a mole in a litre.

These are the kind of double-checks you should do by yourself all the time. Not only to answer questions at university. Professional scientists do it all the time. And even: many determine this way if they have to multiply or divide by a given number - and not through formal transformations of the equations, which are too insecure.

I don't understand 0.08206 atm(L) / mol(K). (K) uses to mean "measured in K", which isn't the case here. Do you mean atm*L/(mol*K)? You must train to care about the units. They help you. Again, not just to answer questions at university. Professional scientists get help from them all the time. It's the basic way to determine whether one shall multiply or divide. When you see a shield stopping factor in cm2/g, the unit tells you what the possibly unknown thing is and how to use it.

Train these two (direct of inverse relationship, and units) as soon as possible. Obviously you miss them and they are vital. I know that most professors neglect to teach them seriously. Whether a student discovers them by himself determines much whether he or she will be fluent in science and technology.

I dislike the expression of R in atm etc.
  • You will need R for other computations, where no atm is present, so better learn 8.3145... (if SI, that is J/K, or in your preferred chemist unit system).
  • The STP convention has changed (or will change) from 1atm to 1bar. But the existing books have not, so double-check every time.
  • Having to insert P≠1 helps remember that P can vary. That's a frequent cause of error, especially mistaking atm and bar. If P≈105 in you unit system, you notice an error more probably.
  • If really you want to memorize a shortcut for the ideal gas law, then learn the 22.45L instead: it helps double-checks and is faster. But beware atm vs bar and 273K vs 298K vs 300K.

Sponsored Links